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EDIT: fixed asymmetrical denominator!

I was following along with a proof of Routh's theorem, and the final expression for the area of the enclosed triangle is $$ 1 - \frac{x}{xz + x + 1} - \frac{y}{xy + y + 1} - \frac{z}{yz + z + 1} $$ (or something extremely similar).

It's supposed to simplify to $$\frac{(xyz - 1)^2}{(xz + x + 1)(xy + y + 1)(yz + z + 1)} $$

However, I've tried doing the simplification by hand, but to no avail.

Using the WolframAlpha[...] function in Mathematica and then doing a step-by-step solution gives a few pages' worth of expansions that are obviously beyond a mortal.

Is there any way to do this that I'm missing?

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  • $\begingroup$ "or something extremely similar"??? Please check your expressions (both of them) since you give them out of context and don't explain where they come from - and they don't have any obvious symmetry properties (as stated) between $x,y,z$ which would be expected for a triangle. Setting $x=y=0$ gives $1-z=1$, which is false unless $z=0$ and you give no reason why that should be so. $\endgroup$ – Mark Bennet Sep 4 '14 at 18:25
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    $\begingroup$ I'm guessing the last fraction is actually $$\frac z{yz+z+1}$$ to make for full symmetry in the expression. $\endgroup$ – abiessu Sep 4 '14 at 19:08
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    $\begingroup$ @MarkBennet yes, there was a typo in the original question, I've fixed it. $\endgroup$ – Soham Chowdhury Sep 5 '14 at 1:43
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It is easy if you fix your starting expression. The last term is $\frac{z}{yz+z+1}$ you changed that $z$ in the denominator to $y$, making it unsymmetrical.

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Setting your two expressions equal to each other and mutliplying by the denominators and simplifying I get:

$$ (y-z)(1+ x z + x^2 y z) = 0$$

Which means in general they are not equal. Or so it would seem.

Added:

Going off the idea of a typo: if you change the $(y z + y + 1)$ to $( y z + z + 1)$ then it simplifies to the stated answer (with the same change).

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  • $\begingroup$ Yes, but how is the simplification done? I've taken a common denominator and everything, but ... the algebra gets too messy. Is there any "elegant" way? $\endgroup$ – Soham Chowdhury Sep 5 '14 at 2:23
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    $\begingroup$ I don't think there is a simpler way to prove it "by hand" than just getting your common denominator and simplifying. Sometimes nothing will save you from a lot of algebra--although they don't usually belabor such situations in text books. $\endgroup$ – amcalde Sep 5 '14 at 3:50
  • $\begingroup$ Computer algebraic programs can help simplify for you but this is not any more elegant, it's just a short cut. $\endgroup$ – amcalde Sep 5 '14 at 3:53
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Starting with $$1 - \frac{x}{xz + x + 1} - \frac{y}{xy + y + 1} - \frac{z}{yz + y + 1}$$ I arrived to $$\frac{x^2 y^2 z^2+x^2 y^2 z-x^2 y z^2-x y z-x z^2+y-z+1}{(x y+y+1) (x z+x+1) (y z+y+1)}$$ Probably typo's somewhere.

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  • $\begingroup$ Yeah, the denominator of the last fraction should be $yz + z + 1$. $\endgroup$ – Soham Chowdhury Sep 5 '14 at 1:44

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